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DENIUS [597]
3 years ago
14

Can you make 24 from the numbers 3, 6, 8 and 2, using all the numbers? You can only use the operations: + - x / ( )

Mathematics
1 answer:
Andre45 [30]3 years ago
4 0

Answer:

you can do 3 x 6 + 8 - 2

Step-by-step explanation:

3 × 6= 18

+8= 26

-2= 24

hope this Helps :)

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2. Write the probability of each event when choosing one card at random from a standard
xeze [42]

Answer:

Step-by-step explanation:

(King of Diamonds) = 1/52

(A Heart) = 1/4

(Red Three) = 1/26

(Red Card) = 1/2

3 0
3 years ago
how much water should be mixed with 6 gallons of 55% solution of sulfuric acid to make a 20% solution of sulfuric acid
nevsk [136]
Hello.

For all dilution problems use the dilution equation: 
<span>C1V1 = C2V2 </span>
<span>20*V1 = 6*55 </span>
<span>V1 = 6*55/20 </span>
<span>V1 = 16.5 gallons </span>
<span>The final volume must be 16.5 gallons </span>
<span>You must add 16.5 - 6 = 10.5 gallons water to be added.
</span>
Have a nice day
6 0
3 years ago
Homework due in 10 minutes!!!
Brut [27]
$3,456 will be given to the band.

Simply multiplying 57,600 and 6% which is 0.06, you get 3,456.
6 0
3 years ago
Read 2 more answers
Please help solve this question​
Rasek [7]

95141 1404 393

Answer:

  • arc BC: 8.55 cm
  • chord BC: 8.03 cm

Step-by-step explanation:

The length of an arc that subtends central angle α will be ...

  s = rα . . . . where α is in radians

The central angle BOC is twice the measure of angle QBC, so is 70°, or 7π/18 radians. So, the length of arc BC is ...

  s = (7 cm)(7π/18) ≈ 8.55 cm . . . arc BC

__

For central angle α and radius r, the chord subtending the arc is ...

  c = 2r·sin(α/2)

  c = 2(7 cm)sin(35°) ≈ 8.03 cm . . . . chord AB

5 0
2 years ago
Prove that the sum of the squares of any two odd numbers is always even
ValentinkaMS [17]

Answer:

proof below

Step-by-step explanation:

Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)

Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;

(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2

= 2(2a^2 + 2a + 2b^2 + 2b + 1)

Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.

6 0
3 years ago
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